Powers and their bases, the order of operations, substituting into an algebraic expression, identifying terms and coefficients, combining like terms, and building an expression from a verbal model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 1 — Equations and Inequalities
Evaluate and Simplify Algebraic Expressions
Objectives
Five outcomes. The third and fourth are where most Algebra 2 arithmetic errors are actually born.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 10-17 — the lesson these objectives are drawn from
Warm-up
Everybody has been burned by this one at least once.
Discussion prompt
Evaluate 2 plus 3 times 4 two different ways: left to right, and using the order of operations. Which answer does a calculator give, and why is that the agreed answer rather than the other one?
Hint: There are only two candidate answers here, and they are quite far apart.
Answer:
\[ 2 + 3 \cdot 4 = 2 + 12 = 14 \quad \text{(order of operations)} \]
\[ (2 + 3) \cdot 4 = 5 \cdot 4 = 20 \quad \text{(left to right)} \]
Fourteen. The order of operations is a convention, not a discovery — it exists so that one string of symbols has exactly one meaning. When you want the other answer you have to say so with brackets, which is exactly what brackets are for.
Concept
An algebraic expression is a set of instructions waiting for numbers. Simplifying rewrites the instructions into a shorter set that always gives the same answer; evaluating finally runs them. Keeping those two apart is what makes long problems tractable.
equivalent expressions — Two expressions that give the same value for every value of their variables. A statement equating two of them, such as eight x plus three x equals eleven x, is called an identity.
Simplify first, then substitute. Doing it the other way around means carrying long arithmetic through every step.
Figure (svg): The power seven cubed labelled with its base, its exponent, and the repeated multiplication it stands for
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 10-12
Section
Section 1
Concept
A power is repeated multiplication of one factor. The exponent counts how many copies of the base are multiplied; it never means multiply by the exponent.
power — An expression formed by repeated multiplication of the same factor. It has a base, the factor being repeated, and an exponent, the number of copies.
\[ 7^3 = 7 \cdot 7 \cdot 7 = 343 \]
An exponent of one is not usually written, so eight to the first power is simply written as eight.
Figure (svg): The power seven cubed labelled with its base, its exponent, and the repeated multiplication it stands for
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 10-10
Picture it
This is the single most reliable way to lose a mark in Chapter 1, and it is pure notation.
Figure (svg): Two columns contrasting negative five raised to the fourth power in brackets with the negative of five to the fourth power
With brackets the base is negative five and four negatives multiply to a positive. Without brackets the base is five, the power is computed first, and the minus sign is applied at the very end.
Worked example
Example 1. Read the base before you compute anything.
\[ \text{Evaluate } (-5)^4 \text{ and } -5^4. \]
Identify the base of the first power
Why: The brackets enclose the minus sign, so the base is negative five and the whole thing is repeated four times.
\[ (-5) (-5) (-5) (-5) \]
Multiply in pairs
Why: Two negatives make a positive, and there are two such pairs, so the result is positive.
\[ 25 \times 25 = 625 \]
Identify the base of the second power
Why: There are no brackets. Powers are evaluated before the minus sign is applied, so the base is five alone.
\[ -(5 \times 5 \times 5 \times 5) \]
Compute the power, then apply the minus
Why: Five to the fourth is 625, and only then does the minus sign attach.
\[ -625 \]
Figure (svg): The solution to Worked example two powers that differ by a bracket shown as a ladder of expressions, one row per algebraic move
\[ (-5)^4 = 625 \qquad -5^4 = -625 \]
Verify: count the negative factors
Why: In the bracketed version there are four negative factors, and an even count of negatives gives a positive product. In the unbracketed version there is exactly one minus sign and it never entered the multiplication, so the sign of the answer is negative. The magnitudes agree, which is the giveaway that the only difference is the bracket.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 10-10
Sorting
Do not compute these. Decide the sign of each, then check.
Sort into buckets
Sort each power by the sign of its value.
Two questions settle every one of these: is the minus inside the brackets, and is the exponent even or odd?
Worked example
The running example from the Order of Operations box, done one step per line.
\[ \text{Evaluate } 1 + 7^2 \cdot (5 - 3). \]
Do the operation inside the grouping symbols
Why: Grouping symbols outrank everything, and there is only one grouping here.
\[ 1 + 7 ^{2} \times 2 \]
Evaluate the power
Why: Powers come before multiplication, which is why the 7 squared is resolved before it meets the 2.
\[ 1 + 49 \times 2 \]
Multiply, working left to right
Why: Multiplication outranks addition, so the 49 and the 2 combine before the 1 joins in.
\[ 1 + 98 \]
Add
Why: Only addition is left.
\[ 99 \]
Figure (svg): The solution to Worked example order of operations, all four steps shown as a ladder of expressions, one row per algebraic move
\[ 1 + 7^2 \cdot (5 - 3) = 99 \]
Verify: re-do it with the steps deliberately in the wrong order
Why: Adding first would give 8 times 4 squared, or 128 — a different number entirely. That the two orders disagree is exactly why the convention exists, and getting 99 by following the four steps in order confirms the convention was applied.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 10-10
Trap
\[ \text{Evaluate } 3y^2 - 4y \text{ when } y = -2. \]
Compute y squared as -2 times 2
Why: The exponent is being read as a factor rather than as a count of factors.
\[ 3(-4) - 4(-2) = -12 + 8 = -4 \quad \text{(wrong)} \]
\[ \text{Evaluate } 3y^2 - 4y \text{ when } y = -2. \]
Substitute with brackets, then square
Why: Squaring means multiplying negative two by itself, which gives positive four.
\[ 3(-2)^2 - 4(-2) = 3(4) + 8 \]
\[ = 12 + 8 = 20 \]
Prediction
Guided Practice items 1 to 3, stacked so the pattern is visible.
Predict first
Which of these three has a different sign from the other two: 6^3, -2^6, or (-2)^6?
Correct: -2^6 is the odd one out — it is the only negative of the three.
\[ 6^3 = 216 \qquad (-2)^6 = 64 \qquad -2^6 = -64 \]
Why: Six cubed is 216, positive because the base is positive. Negative two to the sixth is 64, positive because six negatives pair off evenly. But minus two to the sixth has no brackets, so it is the negative of 64, which is -64. The bracket is the only difference between the last two and it flips the sign.
Ranking
You know these. Putting them in order is the point — most errors are an out-of-order step, not a wrong step.
Put in order
Why: Grouping symbols come first because they are the author's explicit instruction to override the default order. Powers outrank multiplication because a power IS repeated multiplication and has to be resolved into a number before it can be used as a factor. Multiplication outranks addition for the same structural reason. The left-to-right rule only breaks ties inside a level, which matters for division and subtraction because neither is commutative.
Notation
One line from Example 2, with every choice in it worth naming.
Annotate
On: \( -4x^2 - 6x + 11 \;=\; -4(-3)^2 - 6(-3) + 11 \quad \text{when } x = -3 \)
Substituting a negative number without brackets is the second most common source of sign errors in this chapter, after the unbracketed power itself.
Section
Section 2
Concept
Write the expression as a sum, and the parts added together are the terms. A term with a variable part is a variable term; one without is the constant term. The number multiplying a power is its coefficient.
like terms — Terms that have exactly the same variable parts, meaning the same variables raised to the same powers. Constant terms count as like terms with each other.
Writing minus seven as plus negative seven is not pedantry — it is how you stop a minus sign getting attached to the wrong term when you rearrange.
Figure (svg): The expression three x squared plus five x plus negative seven with its terms, coefficients and constant term labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 12-12 — Terms and Coefficients
Picture it
The exponent has to match too, and this is the distinction the textbook flags with an Avoid Errors note.
Figure (svg): Two columns contrasting pairs of like terms with pairs that share a variable but differ in exponent
Three p squared and p use the same letter but different exponents, so they cannot be combined. They are as unlike as three metres and three square metres.
Worked example
Guided Practice 8. Naming the parts before combining them is what makes the combining automatic.
\[ 2 + 5x - 6x^2 + 7x - 3 \]
Rewrite every subtraction as addition of a negative
Why: Terms are the parts ADDED together, so the definition only applies once the expression is a sum.
\[ 2 + 5 x + (-6 x ^{2}) + 7 x + (-3) \]
List the terms
Why: There are five, and the minus signs now travel with the terms they belong to.
\[ 2, 5 x, -6 x ^{2}, 7 x, -3 \]
Name the coefficients and the constant terms
Why: The coefficients are the numbers multiplying the powers; the constants are the terms with no variable at all.
\[ \text{coefficients } 5, -6, 7;\text{ constants } 2\text{ and } -3 \]
Group the like terms
Why: The two x terms are alike; the two constants are alike; the x squared term has no partner.
\[ -6 x ^{2} + (5 x + 7 x) + (2 - 3) \]
Combine each group by adding coefficients
Why: This is the distributive property read backwards, which is why only like terms can be combined at all.
\[ -6 x ^{2} + 12 x - 1 \]
Figure (svg): The solution to Worked example label everything, then simplify shown as a ladder of expressions, one row per algebraic move
\[ 2 + 5x - 6x^2 + 7x - 3 = -6x^2 + 12x - 1 \]
Verify: substitute the same number into both forms
Why: Take x equal to 2. The original gives 2 plus 10 minus 24 plus 14 minus 3, which is -1. The simplified form gives -24 plus 24 minus 1, which is -1 as well. Agreeing at one value is not a proof, but disagreeing at one value would be a disproof, so it is a real check.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 12-12
Discrimination
Do not combine anything. Just decide which pairs are eligible.
Sort into buckets
Sort each pair by whether the two terms can be combined into one.
Worked example
Example 4c. Two brackets, one of them with a minus sign in front — the case worth slowing down for.
\[ 3(y + 2) - 4(y - 7) \]
Distribute the 3 across the first bracket
Why: Every term inside gets multiplied, not just the first one.
\[ 3 y + 6 - 4(y - 7) \]
Distribute negative four, not four, across the second bracket
Why: The minus sign in front belongs to the multiplier. Treating it as a separate subtraction is where this goes wrong.
\[ 3 y + 6 - 4 y + 28 \]
Group the like terms
Why: The y terms together, the constants together.
\[ (3 y - 4 y) + (6 + 28) \]
Combine each group
Why: Three minus four is negative one, and six plus twenty-eight is thirty-four.
\[ -y + 34 \]
Figure (svg): The solution to Worked example distribute, then combine shown as a ladder of expressions, one row per algebraic move
\[ 3(y + 2) - 4(y - 7) = -y + 34 \]
Verify: substitute a value into both forms
Why: Take y equal to 5. The original is 3 times 7 minus 4 times negative 2, which is 21 plus 8, or 29. The simplified form is negative 5 plus 34, which is 29. The sign of the y term is the thing this check is really confirming.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 12-12
Trap
\[ 8(x - 3) - 2(x + 6) \]
Distribute 8, then distribute 2 and subtract only the first product
Why: The minus sign is treated as an operation between brackets rather than as part of the multiplier.
\[ 8x - 24 - 2x + 12 \quad \text{(wrong)} \]
\[ = 6x - 12 \]
\[ 8(x - 3) - 2(x + 6) \]
Distribute negative two across the whole second bracket
Why: The multiplier is negative two, so both terms inside change sign.
\[ 8x - 24 - 2x - 12 \]
\[ = 6x - 36 \]
The x terms happen to agree either way, which is exactly what makes this error survive a quick glance. The constant is what gives it away.
Fill the middle
The grouping step has been rubbed out. Put it back.
Fill in the blanks
5p^2 + p - 2p^2 \;=\; (5p^2 - 2p^2) + p \;=\; 3p^2 + p
Why: Only the two p squared terms are like, so those are the ones that get bracketed together; the lone p is carried along untouched because it has no partner. Writing the grouping line explicitly is what stops the p from being swept into the subtraction, which is the usual failure here.
Error analysis
A student simplifies an expression from the exercise set. The answer is wrong by one sign.
Annotate
On: \( -4y - x + 10x + y \;=\; (-4y + y) + (-x + 10x) \;=\; -3y - 9x \)
When you combine coefficients, do the signed arithmetic on the coefficients alone and attach the variable afterwards.
Two truths and a lie
An identity has to hold for every value of the variable, not just for a convenient one.
Eliminate the wrong options
Two of these hold for every value. Knock those out and keep the one that does not.
Survives elimination: C
Why: The survivor is the false one. Adding five p squared and p treats them as like terms when their exponents differ, so it only accidentally holds when p is 0 or 1. Test p equal to 2: the left side is 20 plus 2, which is 22, and the right side is 24. An identity must hold everywhere, so one failing value is enough to disqualify it.
Section
Section 3
Concept
A verbal model is the sentence version of the expression: each quantity named, each with its unit. Writing it before you write algebra is what stops you multiplying two things that should have been added.
verbal model — A word equation describing a real situation, written before the algebraic expression, with each quantity labelled by its unit.
The unit check is free and it is decisive: dollars per candle times candles gives dollars, so the expression is at least the right kind of thing.
Figure (svg): A verbal model bar showing price per candle times number sold minus expenses, with the algebraic expression beneath it
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 11-13
Picture it
Example 3: candles at three dollars each, and a hundred and twenty dollars of costs.
Figure (svg): A verbal model bar showing price per candle times number sold minus expenses, with the algebraic expression beneath it
Every box in the top row becomes one symbol in the bottom row. Nothing appears in the algebra that was not named in the model first.
Worked example
Example 3, both parts. Notice the expression is built before any number is substituted.
\[ \text{Candles sell for } \$3 \text{ each; the booth and materials cost } \$120. \text{ Find the profit on 75 candles.} \]
Write the verbal model: profit is income minus expenses
Why: Naming the relationship in words fixes whether the 120 is subtracted or added before any symbols appear.
Attach units to each quantity
Why: Dollars per candle times candles gives dollars, and the expenses are already dollars, so the subtraction is legal.
Write the algebraic expression
Why: This is the answer to the first part of the question, and it works for any number of candles.
\[ 3 c - 120 \]
Substitute 75 for c and evaluate
Why: Only now does a specific number enter, which keeps the arithmetic to one line.
\[ 3(75) - 120 = 225 - 120 \]
Figure (svg): The solution to Worked example profit from a craft fair shown as a ladder of expressions, one row per algebraic move
\[ 3c - 120 \quad \text{and at } c = 75, \; \text{profit} = \$105 \]
Verify: check the break-even point
Why: Setting the expression to zero gives 3c equal to 120, so c is 40 — forty candles just covers the costs. Seventy-five candles is thirty-five past break-even, and thirty-five times three dollars is 105 dollars, matching the answer by an entirely different route.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 11-11
Step zero
A truck used 28 gallons and travelled 428 miles, at 16 miles per gallon on the highway and 12 in the city.
Discussion prompt
Do not solve this. In plain English, say what one letter you would introduce, what the other quantity then has to be in terms of it, and which two products you would add. Say what unit each product carries.
Hint: There are two unknowns but only one letter is needed, because the two gallon counts must total 28.
Answer:
Let g be the gallons used in the city. Then the highway gallons must be 28 minus g, because the two counts total twenty-eight.
\[ 428 = 12g + 16(28 - g) \]
Each product is miles per gallon times gallons, which gives miles — so adding them to a total distance in miles is legitimate. Naming the second quantity in terms of the first is the move that turns two unknowns into one.
Worked example
Example 5. Fifteen prints, some large and some small — and the second quantity is defined by the first.
\[ \text{15 prints: large cost } \$0.80, \text{ small cost } \$0.20. \text{ Find the cost when } n = 5 \text{ are large.} \]
Express the number of small prints in terms of n
Why: The total is fixed at fifteen, so once n are large the remaining count is forced. This is the step the problem is really testing.
\[ \text{small prints } = 15 - n \]
Write the verbal model as price times count, twice, added
Why: Two categories of print, each contributing dollars, so the two contributions add.
\[ 0.8 n + 0.2(15 - n) \]
Distribute the 0.2 across its bracket
Why: The bracket has to be opened before like terms can be seen.
\[ 0.8 n + 3 - 0.2 n \]
Group and combine the n terms
Why: Eight tenths minus two tenths is six tenths.
\[ 0.6 n + 3 \]
Substitute 5 for n
Why: The simplified form makes this one multiplication and one addition instead of two multiplications and a subtraction.
\[ 0.6(5) + 3 = 3 + 3 \]
Figure (svg): The solution to Worked example simplify the model before evaluating shown as a ladder of expressions, one row per algebraic move
\[ 0.8n + 0.2(15 - n) = 0.6n + 3, \quad \text{and at } n = 5, \; \text{cost} = \$6 \]
Verify: count the prints directly
Why: Five large prints at eighty cents is four dollars, and ten small prints at twenty cents is two dollars, for six dollars total. The direct count agrees with the simplified expression, confirming both the algebra and the fifteen-minus-n step.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 13-13
Error analysis
A student sets up the print-cost problem and gets an expression that cannot be right — and the units say so before the arithmetic does.
Annotate
On: \( \text{cost} = 0.8n + 0.2 \cdot 15 - n \)
A dropped bracket almost always shows up as a unit mismatch, which is why writing the verbal model in units is worth the extra line.
Matching
Each situation becomes exactly one of these expressions.
Match the pairs
Why: Two structural clues sort all four. A fixed starting amount that is then adjusted becomes a constant plus or minus a rate times a variable, which covers the third and fourth. A quantity split into two categories becomes two products added, with the second category written as the total minus the first, which is the second one. And a straightforward income-minus-costs becomes a product minus a constant.
Pattern
Height of a paramotorist, in feet, at each minute of a descent.
Step through it
What is the height after 7 minutes, and what does the model predict for the moment of landing?
At seven minutes the height is 2000 minus 1750, which is 250 feet. The model reaches zero at t equal to 8, so it only describes the descent up to that point.
Reverse engineer
Here is a simplified expression. Something real produced it.
Fill in the blanks
\text0.6 = ___\,n + 3
Why: The blank is the coefficient of n, and it is the extra cost of upgrading one print from small to large: eighty cents instead of twenty. The constant 3 is what you would pay if every one of the fifteen prints were small, which is fifteen times twenty cents. Reading a simplified expression back into its story is how you check that a simplification did not lose the meaning.
Section
Section 4
Concept
Substitution replaces a letter with a number. If that number is negative, the brackets are not optional — they are what keeps the minus sign attached to the value rather than floating into the operation next to it.
\[ 3y^2 - 4y \;\text{ at }\; y = -2 \;\longrightarrow\; 3(-2)^2 - 4(-2) \]
Every substituted value gets brackets, even a positive one, so that the habit does not have to be remembered under pressure.
Figure (svg): Two columns contrasting negative five raised to the fourth power in brackets with the negative of five to the fourth power
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 11-11
Picture it
The order of operations is the schedule that substitution has to obey.
Figure (svg): Four numbered steps of the order of operations, each with the running example one plus seven squared times the quantity five minus three
Substitution happens first of all, before step one. After that the expression is purely numerical and the four steps take over.
Worked example
Example 2. Four separate sign decisions in one short expression.
\[ \text{Evaluate } -4x^2 - 6x + 11 \text{ when } x = -3. \]
Substitute negative three for every x, in brackets
Why: Two occurrences, two brackets. Writing this line separately is what makes the rest mechanical.
\[ -4(-3) ^{2} - 6(-3) + 11 \]
Evaluate the power first
Why: Negative three squared is positive nine, because the brackets put the minus sign inside the power.
\[ -4(9) - 6(-3) + 11 \]
Do the two multiplications
Why: Negative four times nine is negative thirty-six. Negative six times negative three is positive eighteen.
\[ -36 + 18 + 11 \]
Add left to right
Why: Negative thirty-six plus eighteen is negative eighteen; plus eleven is negative seven.
\[ -7 \]
Figure (svg): The solution to Worked example evaluate with a negative value shown as a ladder of expressions, one row per algebraic move
\[ -4x^2 - 6x + 11 = -7 \quad \text{when } x = -3 \]
Verify: re-evaluate with the terms in a different order
Why: Adding the positives first gives eighteen plus eleven, which is twenty-nine, and twenty-nine minus thirty-six is negative seven. Reaching the same value by regrouping confirms the arithmetic without repeating the same keystrokes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 11-11
Elimination
The expression is 3y squared minus 4y, evaluated at y equal to negative two.
Eliminate the wrong options
Which value is correct?
Survives elimination: A
Why: Squaring negative two gives positive four, so the first term is three times four, which is twelve. The second term is minus four times negative two, which is positive eight. Twelve plus eight is twenty. Both terms end up positive, which is the surprise worth remembering about even powers of negative inputs.
Worked example
Guided Practice 4. The bracket has to be resolved before the multiplication outside it.
\[ \text{Evaluate } 5x(x - 2) \text{ when } x = 6. \]
Substitute 6 for both occurrences of x
Why: There are two, one outside the bracket and one inside, and both must change.
\[ 5(6) ((6) - 2) \]
Do the operation inside the grouping symbols first
Why: Six minus two is four, and grouping symbols outrank multiplication.
\[ 5(6) (4) \]
Multiply left to right
Why: Five times six is thirty; thirty times four is one hundred and twenty.
\[ 120 \]
Figure (svg): The solution to Worked example a variable inside a bracket shown as a ladder of expressions, one row per algebraic move
\[ 5x(x - 2) = 120 \quad \text{when } x = 6 \]
Verify: expand the expression first, then substitute
Why: Distributing gives 5x squared minus 10x. At x equal to 6 that is 5 times 36 minus 60, which is 180 minus 60, or 120. Getting the same answer from the expanded form confirms both the substitution and the distribution.
Trap
\[ \text{Evaluate } x^2 + 5x \text{ when } x = -4. \]
Write the substitution without brackets
Why: The minus sign is left loose, so it is no longer clear what it belongs to.
\[ -4^2 + 5 \cdot -4 = -16 - 20 = -36 \quad \text{(wrong)} \]
\[ \text{Evaluate } x^2 + 5x \text{ when } x = -4. \]
Bracket every substituted value
Why: The brackets say that the whole of negative four is the thing being squared and the thing being multiplied.
\[ (-4)^2 + 5(-4) = 16 - 20 \]
\[ = -4 \]
The two answers differ by thirty-two, all of it from one missing pair of brackets in the very first line.
Explain it to yourself
Two substitution lines, differing only in punctuation.
\[ -4(-3)^2 \qquad \text{versus} \qquad -4-3^2 \]
Discussion prompt
Explain, in your own words, what each of these two lines instructs you to do and why they give different answers. Then say what would have to be written to make the second line mean the first.
Hint: Ask what the base of the power is in each case, and what the minus signs are attached to.
Answer:
\[ -4(-3)^2 = -4 \cdot 9 = -36 \]
\[ -4 - 3^2 = -4 - 9 = -13 \]
In the first line the brackets make negative three the base and the multiplication by negative four happens afterwards. In the second there is no multiplication at all — the missing bracket turned a product into a subtraction. Brackets carry the multiplication, which is why dropping them changes the operation, not just the sign.
Commit first
Rate your confidence honestly on this one.
Predict first
For which values of n is (-2)^n negative?
Correct: Odd n only.
\[ (-2)^2 = 4, \; (-2)^3 = -8, \; (-2)^4 = 16, \; (-2)^5 = -32 \]
Why: Each factor contributes one minus sign, and minus signs cancel in pairs. An even count pairs off completely and leaves a positive result; an odd count leaves exactly one unpaired minus sign, so the result is negative. The magnitude of n has nothing to do with it, only whether it is even or odd, which is why 2 to the 100th and 2 to the 101st differ in sign at all.
Edge cases
The rules for powers have to keep working at the ends of their range.
Discussion prompt
What does an exponent of 1 mean, and what would an exponent of 0 have to mean if the pattern of dividing by the base each time you step down is to keep holding? Try it with base 3 and see where it forces you.
Hint: Write 3 to the fourth, third, second and first, and look at what happens between consecutive entries.
Answer:
\[ 3^4 = 81, \; 3^3 = 27, \; 3^2 = 9, \; 3^1 = 3 \]
Each step down divides by three. Continuing the pattern one more step forces three to the zero to be three divided by three, which is 1 — not zero, which is the usual guess.
This lesson only needs the exponent-of-one case, where the exponent is not written at all. But the pattern that forces it is the same one that will define negative and fractional exponents in Chapters 5 and 6, so it is worth seeing now.
Section
Section 5
Concept
Given an expression and a value, simplifying before substituting almost always saves work and always reduces the number of places an arithmetic slip can hide. The two forms are equivalent, so the answer cannot differ.
identity — A statement equating two equivalent expressions, such as eight x plus three x equals eleven x. It holds for every value of the variable, not just for some.
This ordering is a habit worth building now: in later chapters the expressions get long enough that substituting first becomes genuinely impractical.
Figure (svg): The expression three x squared plus five x plus negative seven with its terms, coefficients and constant term labelled
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 12-13
Picture it
Simplifying is done under the same schedule as evaluating.
Figure (svg): Four numbered steps of the order of operations, each with the running example one plus seven squared times the quantity five minus three
The only difference is that when a step involves an unknown, you leave it as a term instead of collapsing it to a number.
Worked example
Guided Practice 13, and then a value substituted, so the saving is visible.
\[ \text{Simplify } 8(x - 3) - 2(x + 6), \text{ then evaluate at } x = 10. \]
Distribute 8 and negative 2 across their brackets
Why: The second multiplier is negative two, so both terms in that bracket change sign.
\[ 8 x - 24 - 2 x - 12 \]
Group like terms
Why: Two x terms and two constants.
\[ (8 x - 2 x) + (-24 - 12) \]
Combine each group
Why: Eight minus two is six; negative twenty-four minus twelve is negative thirty-six.
\[ 6 x - 36 \]
Substitute 10 for x
Why: One multiplication and one subtraction, instead of two brackets and two products.
\[ 6(10) - 36 = 60 - 36 \]
Figure (svg): The solution to Worked example simplify then evaluate shown as a ladder of expressions, one row per algebraic move
\[ 8(x - 3) - 2(x + 6) = 6x - 36 \qquad \text{and at } x = 10, \; 24 \]
Verify: evaluate the original expression at the same value
Why: The original at x equal to 10 is 8 times 7 minus 2 times 16, which is 56 minus 32, or 24. Both forms give 24, so the simplification preserved the value — which is what equivalent means.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 12-12
Comparison
Fill the blanks. Both columns must land on the same value, which is the point.
Comparison matrix
| Step | Substitute first | Simplify first |
|---|---|---|
| Start | 8(x - 3) - 2(x + 6) at x = 10 | 8(x - 3) - 2(x + 6) |
| Next | 8(7) - 2(16) | 8x - 24 - 2x - 12 |
| Then | 56 - 32 | 6x - 36 |
| Finish | 24 | 6(10) - 36 = 24 |
Both are correct. The right-hand column costs one extra line now and saves that line every subsequent time the expression is evaluated, which is why it is the habit worth having.
Worked example
Example 5 again, this time treated purely as an algebra exercise so the structure stands out.
\[ \text{Simplify } 0.8n + 0.2(15 - n) \text{ and interpret both parts of the result.} \]
Distribute the 0.2
Why: Two tenths times fifteen is three; two tenths times negative n is negative two tenths n.
\[ 0.8 n + 3 - 0.2 n \]
Group the n terms
Why: The constant has no partner, so it comes along unchanged.
\[ (0.8 n - 0.2 n) + 3 \]
Combine
Why: Eight tenths minus two tenths is six tenths.
\[ 0.6 n + 3 \]
Interpret each part
Why: The constant is the cost if every print were small; the coefficient is what each upgrade to large costs on top of that.
\[ 0.6 n + 3 \]
Figure (svg): The solution to Worked example a model with a bracket and a rate shown as a ladder of expressions, one row per algebraic move
\[ 0.8n + 0.2(15 - n) = 0.6n + 3 \]
Verify: check both endpoints
Why: At n equal to 0 the simplified form gives 3 dollars, which is fifteen small prints at twenty cents — correct. At n equal to 15 it gives 9 plus 3, or 12 dollars, which is fifteen large prints at eighty cents — also correct. Both extremes matching is a much stronger check than one middle value.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 13-13
Error analysis
A student simplifies, then evaluates at x equal to 4. The final number is wrong, and exactly one line is responsible.
Annotate
On: \( \begin{aligned} 5(x - 2) - 3(x + 1) &= 5x - 10 - 3x + 3 \\ &= 2x - 7 \\ \text{at } x = 4: \quad &= 2(4) - 7 = 1 \end{aligned} \)
Note that the error survived the second and third lines untouched. Checking only the last step would never have found it, which is the argument for substituting into the ORIGINAL expression when you verify.
Explain it
They keep asking why they cannot add 5p squared and p.
Discussion prompt
Without using the phrase like terms, explain to them why five p squared and p cannot be added into one term. Give them a concrete number to test it with, and give them an analogy from outside mathematics.
Hint: Try p equal to 2 or 3 and compare the honest answer with what they expect.
Answer:
At p equal to 3, five p squared is 45 and p is 3, giving 48. Their proposed six p squared is 54. The gap grows as p grows, so it is not a rounding issue — the two expressions are simply different functions.
The analogy that lands: five square metres and three metres. Both mention metres, but one is an area and one is a length, and adding them produces nothing. The exponent is what makes them different kinds of quantity.
Definition probe
The definition of simplified has two clauses: no grouping symbols left, and all like terms combined.
Sort into buckets
Sort each expression by what is still wrong with it.
Counterexample
A classmate proposes a shortcut.
\[ a(b + c) = ab + c \]
Discussion prompt
This says you only need to multiply the first term inside the bracket. Find numbers that break it, then say in one sentence what the correct statement is and why the shortcut is tempting.
Hint: Almost any numbers will break it. Pick ones where c is not zero.
Answer:
\[ 2(3 + 5) = 2 \cdot 8 = 16 \qquad \text{but} \qquad 2 \cdot 3 + 5 = 11 \]
The correct statement is that the factor is handed to every term inside: a times b plus a times c. The shortcut is tempting because it is true whenever c is zero, and because it matches how the eye reads left to right and then stops.
Comparison
Fill in the blanks. The question wording tells you which column you are in.
Comparison matrix
| The question says | What you produce | Is a number the answer? |
|---|---|---|
| Evaluate 6 cubed | a single number | yes |
| Simplify 8x + 3x | a shorter equivalent expression | no |
| Evaluate 3y^2 - 4y when y = -2 | a single number | yes |
| Write an expression for the profit | an expression in the variable | no |
| Simplify, then find the cost when n = 5 | expression first, then a number | yes, at the end |
The last row is the shape of most real problems, and it is the one where doing the steps in the other order costs you the most arithmetic.
Pattern
One routine covers evaluating, simplifying and modelling.
Steps two and five are where sign errors are born, and both are cured by writing one extra line rather than by being more careful.
OpenStax Algebra and Trigonometry 2e, §1.1 Real Numbers: Algebra Essentials §1.1
Check
Powers and brackets. Read the base before you compute.
Check your understanding
Which pair of values is correct for (-3)^4 and -3^4, in that order?
Answer: A
Why: With brackets the base is negative three, and four negative factors pair off to give positive 81. Without brackets the base is three, so the power is 81 and the minus sign is applied afterwards, giving -81.
Check
Combining like terms with a bracket in front of a minus sign.
Check your understanding
Simplify 3(y + 2) - 4(y - 7).
Answer: A
Why: Distributing gives 3y + 6, then negative four times y is -4y and negative four times -7 is +28. Collecting: 3y - 4y is -y, and 6 + 28 is 34.
Check
Substitution with a negative value. Write the substitution line out.
Check your understanding
Evaluate -4x^2 - 6x + 11 when x = -3.
Answer: A
Why: Negative three squared is positive nine, so the first term is -4 times 9, which is -36. The second term is -6 times -3, which is +18. Then -36 + 18 + 11 gives -7.
Real world
A phone plan charges 25 dollars a month plus 12 cents for each minute over the 500 included. You used m minutes, where m is more than 500.
Discussion prompt
Write an expression for the monthly bill, simplify it, and then say what each of the two numbers in your simplified form means in the real situation. What does the expression predict for m equal to 500, and is that prediction correct?
Hint: The number of billable minutes is not m.
Answer:
\[ \text{bill} = 25 + 0.12(m - 500) \]
\[ = 25 + 0.12m - 60 = 0.12m - 35 \]
The 0.12 is the cost per minute once you are paying by the minute; the -35 is what falls out of the algebra, not a real discount. At m equal to 500 it gives 60 minus 35, which is 25 dollars — correct, because no overage has been used yet.
But the simplified form is only valid for m at least 500. Below that the bill is a flat 25 dollars, and the expression would wrongly predict less. Simplifying never changes an expression's value, but it can hide the domain the model was built for.
Commit first
Answer, then rate your confidence. This one catches a lot of confident people.
Predict first
Is -x always a negative number?
Correct: No — it is negative only when x is positive.
\[ x = 6 \Rightarrow -x = -6 \qquad x = -6 \Rightarrow -x = 6 \qquad x = 0 \Rightarrow -x = 0 \]
Why: The expression -x means the opposite of x, not a negative number. If x is -6 then -x is +6, which is positive. If x is 0 then -x is 0, which is neither. Reading the minus sign as an instruction to flip, rather than as a label meaning negative, is the distinction that keeps sign work honest for the rest of the course.
Explain it
They can already do arithmetic but have never had to simplify before substituting.
Discussion prompt
In four sentences or fewer, explain why it is usually better to simplify an expression before putting numbers in. Give them the one situation where it makes no difference, and the one habit that prevents most sign errors.
Hint: Think about how many times each form has to be evaluated.
Answer:
A usable answer: simplifying shortens the recipe, so every later evaluation is quicker and has fewer places to slip. It makes no difference when you only ever need one value — the two forms must agree, since they are equivalent.
The habit is brackets: put every substituted value in brackets, even the positive ones, so the minus signs stay attached to the numbers they belong to.
Exit ticket
Name the weakest spot. That is tonight's ten minutes.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is the one to work on, and each has a one-line fix.
Why: For powers, read the base out loud before computing. For distributing, write the multiplier with its sign attached before you start. For like terms, compare the variable parts including exponents, not just the letters. For word problems, write the verbal model with units before any symbols. Pick yours and do five of that kind rather than twenty mixed ones.
Connect it up
Paper, not screen. This is the part that sticks.
Draw it
Across the top of a page write one power with brackets and one without, and beside each write its value and which number is the base. Down the left side write the four steps of the order of operations, and beside each write one line of the expression one plus seven squared times the quantity five minus three, showing what that step changed. In the middle of the page write a five-term expression of your own, circle the terms, box the coefficients, and star the constant term. At the bottom, invent a small situation with a fixed cost and a per-item cost, write its verbal model in units, turn it into an expression, simplify it, and then evaluate it at one value. Finally, draw an arrow from your simplified expression back to the situation and write what each number in it means.
If the last arrow is hard to write, the expression is right but the model was never really understood — go back to Section 3.
Recap
Five things, and the last one is the one that turns word problems from guesswork into procedure.
| If you see | The move is |
|---|---|
| A minus sign next to a power | Ask whether it is inside the brackets |
| A bracket with a number in front | Distribute, sign included |
| Two terms with the same letter | Check the exponents before combining |
| A negative value being substituted | Wrap it in brackets first |
| A word problem | Verbal model with units, then symbols |
Lesson 1.3 takes these same expressions and sets two of them equal, which is all an equation is.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.2 Evaluate and Simplify Algebraic Expressions §1.2, pp. 10-17 — everything on these slides traces back here
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